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In a 110 V DC chopper drive using the CLC scheme, the maximum possible value of the accelerating current is 300 A. The lower limit of the current pulsation is 140 A. What is the maximum limit of current pulsation?
Current Limit Control of Chopper (CLC):
In this strategy, the chopper is turned on when the output current i_{0} equals a preset value I_{2}.
The chopper is kept on till i_{0} increases to another preset value I_{1}.
The chopper is turned off when i_{0} equals I_{1} and is kept off till the current i_{0} decays to I_{2}.
The current i_{0} is thus limited to a maximum of I_{1} and a minimum of I_{2} as shown in Figure.
The chopping frequency and the pulse width are dependent on the load parameters.
The advantage of current limit control is that the ripple (I_{1}  I_{2}), in the load current i_{0} can be adjusted to a small, predetermined value for a wide range of loads.
Application:
Given,
The maximum possible value of the accelerating current (I_{2}) = 300 A
Current pulsation or current ripple (ΔI) = 140 A
The waveform can be drawn as,
From the waveform,
The maximum limit of current pulsation (I_{2}) = I_{1} + ΔI = 300 + 140 = 440 A
In the LC circuit shown in the figure initial current through the inductor is zero, initial voltage across the capacitor is 100 V. Switch S is closed at t = 0 sec. The current through the circuit is
Concept:
In a LC circuit, the current flows in the circuit is given by,
Where, V_{S} is supply voltage in V
C is capacitance in F
L is inductance in H
ω_{0} is resonant frequency in rad/sec
Calculation:
From the given circuit, V_{S} = 100 V
C = 1 μF
L = 20 mH
In the circuit shown in figure find the circuit turn off time is ______ μs
In voltage commutation circuit
Circuit turn off time
for highly inductive load
t_{m} = RC ln 2  for resistive load
∴ t_{m} = 15 × 8 × 10^{6} ln 2 = 83.18 μs
Consider the circuit shown below:
The effective on period of the chopper if V =230 V,I_{o} = 60 A,C = 55 μF,T_{on} = 800 μs is:
Concept:
Mode 1: T_{1} will remain ON and conduct to load and diode T_{A} will remain OFF
Mode 2: T_{A} will be turned ON; the capacitor current will now flow in reverse direction & T1 stops conducting.
Therefore, the effective time period of the chopper is
Commutation time: It is the time taken to disconnect the load from the supply after the main thyristor is turned OFF.
Calculation:
Effective on period
= (0.8 + 0.42) × 10^{3}
T_{ON}’ = 1.22 ms
What value of capacitor (in μF) is required to forces commutate a thyristor with a turn off time of 20 μs with a 96 V battery and a full load current of 100 A.
The size of capacitor enquired for commutation is
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